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SNIIX18

  • 2 years ago

If the period of a pendulum is 2.0 seconds and the acceleration due to gravity is 9.8 meters/second^2, What is the length of the cord on the pendulum? A)1.0 meters B)0.67 meters C)1.3 meters D)3.5 meters

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  1. SNIIX18
    • 2 years ago
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    can you help me

  2. theEric
    • 2 years ago
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    Are you speaking of a simple pendulum?

  3. SNIIX18
    • 2 years ago
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    YES

  4. theEric
    • 2 years ago
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    Then you'll want the equation that uses math to describe the relationships between different physical quantities. I think I've show it to you before. I found it at http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html It is:\[T=2\pi\sqrt{\frac{L}{g}}\] \(T\) is the period. \(L\) is the length of the chord. \(g\) is the acceleration caused by gravity. You want to solve for \(L\).

  5. SNIIX18
    • 2 years ago
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    OKAY HOLD ON

  6. SNIIX18
    • 2 years ago
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    I GOT O.67 IS THAT RIGHT

  7. theEric
    • 2 years ago
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    It's not what I got! How did you solve for \(L\)?

  8. SNIIX18
    • 2 years ago
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    I squared 2 then i divided 2.0 by 9.8

  9. theEric
    • 2 years ago
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    I'm confused... Can you show me what you got? You squared the 2?

  10. SNIIX18
    • 2 years ago
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    now im confused cause your confused

  11. theEric
    • 2 years ago
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    Haha, you start with \[T=2\pi\sqrt{\frac{L}{g}}\]Now, what did you square?

  12. SNIIX18
    • 2 years ago
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    owwww okay i see what i did wrong i was suppost to pi the 2 instead of square

  13. theEric
    • 2 years ago
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    Hmmm?

  14. theEric
    • 2 years ago
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    I would divide the equation by \(2\pi\), square the equation, and multiply the equation by \(g\), if you know what I mean. Then you have \(L\) by itself after applicable cancelations, if you know what I mean.

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